Theorem 6: the Binomial Theorem

Have you ever noticed a similarity between the first few powers of 11 and the first few rows of Pascal‘s triangle? I’ll write them out, so that you can compare them. Here are the first few powers of 11.

.

And here are the first few rows of Pascal’s triangle.

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

Notice anything interesting? I’d like to explore the link in this post.

Perhaps the first thing is to see whether the pattern continues. The next row of Pascal’s triangle is

1 5 10 10 5 1,

and is 161051. So something seems to go wrong. Let’s try to understand this, because there does seem to be something curious going on here. It would be great if we could understand both why the first few rows are the same, and why the sixth rows are different.

Understanding the links

How do you calculate the powers of 11? Here’s how I found , having found that is 1331. I want to multiply 1331 by 11. That’s the same as multiplying 1331 by 10 (which is easy), and then adding 1331. (If you do it using long multiplication, that’s exactly what you’re doing.) So I want to add 13310 and 1331.

13310

+ 1331

14641

I’m adding 1331 to “1331 shifted one place to the left with a 0 after it”. So a digit in the answer is the sum of two adjacent digits in 1331 (where we can imagine a 0 before and after 1331: 013310). But that’s exactly how Pascal’s triangle is generated: each digit is the sum of the two adjacent digits above it!

That appears to explain the link, but why does the connection break down when we get to the sixth row? Let’s see what the calculation of looks like. We know that is 14641, so we want to add 10 times that (146410) to 14641.

146410

+ 14641

161051

Ah. So the problem occurs because of “carrying”: becomes “0 carry 1”, rather than 10. So there’s a sense in which it feels as though the pattern would continue, if it weren’t for this pesky business of place value!

What next? I’d like to show you how this relates to something a bit more general.

A more general result

When we looked at powers of 11, we decomposed 11 as , and that helped us to understand what was going on. I’d like to think now about powers of the more general expression . Let’s write out the first few powers. (I’m going to skip the intermediate calculations that I might have used to get these — you might like to do the calculations to convince yourself.)

.

Notice anything about the numbers appearing here? (They’re called the coefficients. For example, the coefficient of in is 2, and the coefficient of is 1.) They’re exactly our old friends from Pascal’s triangle again!

Can we explain this? The good news is that it’s just the same sort of idea as last time. Let’s think about finding , once we know that . We need to multiply by . That is, we need to multiply by , and add it to multiplied by . It’s perhaps not quite as obvious how to write it out this time, but I’ll have a go.

.

Now we need to add these. I’ll try to align them helpfully to make that easier (although I’ve struggled to get LaTeX to cooperate, so it’s not as pretty as I’d like!).

It’s just the same idea as before: adding pairs of adjacent coefficients!

Theorem (the binomial theorem) For any real numbers and , and any non-negative whole number , the coefficients of the various terms in are the numbers of the row of Pascal’s triangle. More precisely, the terms are of the form , where is a whole number between 0 and (inclusive), and is the entry in the row of Pascal’s triangle.

The numbers have a special name: they are called the binomial coefficients! They have many interesting properties. For example, is the number of ways to choose objects from a collection of objects (where the order of the objects doesn’t matter). In how many ways can you choose three of the seven dwarfs? Picking Sleepy, Happy and Sneezy is the same as picking Happy, Sneezy and Sleepy: that’s what I mean when I say that the order doesn’t matter. The answer is that there are ways. This is why is read as “ choose “.

I could say lots more about the binomial coefficients and their interesting properties (including a formula that means that one doesn’t have to write out Pascal’s triangle each time!), but that would make this post too long, so I’ll resist temptation and give some suggestions for further reading instead.

It turns out that there’s a very nice justification of the binomial theorem using the property that is the number of ways to choose objects from (where order is unimportant). I’ll let you think about what that justification is, but here’s a (quite large) hint. Think about as , with bracketed factors. Each term in the answer involves an or a from each bracket. How many times will you get ? What about ?

It turns out that it’s possible to give a more general version of the binomial theorem (sometimes attributed to Newton), in which the exponent doesn’t have to be a non-negative integer. Although this version is both useful and interesting, I shan’t go into it here.

Why have I chosen to write about the binomial theorem? The theorem itself is jolly useful: I can tell you what the coefficient of is in the expansion of , without having to multiply out the whole thing. But for me perhaps even more interesting are the links with Pascal’s triangle and numbers of combinations, and the various properties of the binomial coefficients.

Further reading

Wikipedia has a fairly extensive discussion of the binomial theorem, including other proofs and the more general version I mentioned above. It also has quite a lot about Pascal’s triangle.